Locally principal Ideal in a Noetherian semilocal domain

Question

Let $R$ be a commutative Noetherian domain with only finitely many maximal ideals $\mathfrak m_1,...,\mathfrak m_n$. If $I$ is an ideal of $R$ such that $I_{\mathfrak m_i}$ is a principal ideal of $R_{\mathfrak m_i}$ for every maximal ideal $\mathfrak m_i$ , then is it true that $I$ itself is a principal ideal?

Answer 1

Yes, this is true. Since $R$ is Noetherian, $I$ is finitely generated, and hence finitely presented. The condition that $I_{\mathfrak{m}_{i}}$ is a principal ideal of $R_{\mathfrak{m}_{i}}$ for each $i$ implies that $I$ is a projective $R$-module. (See Bourbaki's "Commutative Algebra", II.5.2 Theorem 1.)

In fact, $I$ has constant rank $1$, since $R$ is a domain, and the rank function $\mathrm{Spec}(R) \to \mathbb{Z}$ sending $\mathfrak{p}$ to $\mathrm{dim}_{R_{\mathfrak{p}}}(I_{\mathfrak{p}})$ is locally constant. (I'll note that one does not need to use the fact that $R$ is a domain here - Theorem 2 of II.5.3 of Bourbaki's "Commutative Algebra" settles this in greater generality.) But $R$ is semi-local by assumption, which shows that $I$ is principal by Proposition 5 of II.5.3 of Bourbaki. (I also wrote about this here in a previous question you asked.)